3.1.82 \(\int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log (\frac {x^2}{3})+x \log (1-x^2)}{\log (\frac {x^2}{3})-\log (1-x^2)}} (2 e^x-2 x+(-1-2 x-x^2) \log ^2(\frac {x^2}{3})+(-x+x^2+2 x^3+e^x (-x^2-x^3)) \log (1-x^2)+(-1-2 x-x^2) \log ^2(1-x^2)+\log (\frac {x^2}{3}) (x-x^2-2 x^3+e^x (x^2+x^3)+(2+4 x+2 x^2) \log (1-x^2)))}{(x^2+x^3) \log ^2(\frac {x^2}{3})+(-2 x^2-2 x^3) \log (\frac {x^2}{3}) \log (1-x^2)+(x^2+x^3) \log ^2(1-x^2)} \, dx\)

Optimal. Leaf size=44 \[ \frac {e^{-x+\frac {(1-x) \left (-e^x+x\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}}}{x} \]

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Rubi [F]  time = 180.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^x*(-1 + x) + x - x^2 - x*Log[x^2/3] + x*Log[1 - x^2])/(Log[x^2/3] - Log[1 - x^2]))*(2*E^x - 2*x + (
-1 - 2*x - x^2)*Log[x^2/3]^2 + (-x + x^2 + 2*x^3 + E^x*(-x^2 - x^3))*Log[1 - x^2] + (-1 - 2*x - x^2)*Log[1 - x
^2]^2 + Log[x^2/3]*(x - x^2 - 2*x^3 + E^x*(x^2 + x^3) + (2 + 4*x + 2*x^2)*Log[1 - x^2])))/((x^2 + x^3)*Log[x^2
/3]^2 + (-2*x^2 - 2*x^3)*Log[x^2/3]*Log[1 - x^2] + (x^2 + x^3)*Log[1 - x^2]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F]  time = 39.51, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {e^x (-1+x)+x-x^2-x \log \left (\frac {x^2}{3}\right )+x \log \left (1-x^2\right )}{\log \left (\frac {x^2}{3}\right )-\log \left (1-x^2\right )}} \left (2 e^x-2 x+\left (-1-2 x-x^2\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-x+x^2+2 x^3+e^x \left (-x^2-x^3\right )\right ) \log \left (1-x^2\right )+\left (-1-2 x-x^2\right ) \log ^2\left (1-x^2\right )+\log \left (\frac {x^2}{3}\right ) \left (x-x^2-2 x^3+e^x \left (x^2+x^3\right )+\left (2+4 x+2 x^2\right ) \log \left (1-x^2\right )\right )\right )}{\left (x^2+x^3\right ) \log ^2\left (\frac {x^2}{3}\right )+\left (-2 x^2-2 x^3\right ) \log \left (\frac {x^2}{3}\right ) \log \left (1-x^2\right )+\left (x^2+x^3\right ) \log ^2\left (1-x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((E^x*(-1 + x) + x - x^2 - x*Log[x^2/3] + x*Log[1 - x^2])/(Log[x^2/3] - Log[1 - x^2]))*(2*E^x - 2
*x + (-1 - 2*x - x^2)*Log[x^2/3]^2 + (-x + x^2 + 2*x^3 + E^x*(-x^2 - x^3))*Log[1 - x^2] + (-1 - 2*x - x^2)*Log
[1 - x^2]^2 + Log[x^2/3]*(x - x^2 - 2*x^3 + E^x*(x^2 + x^3) + (2 + 4*x + 2*x^2)*Log[1 - x^2])))/((x^2 + x^3)*L
og[x^2/3]^2 + (-2*x^2 - 2*x^3)*Log[x^2/3]*Log[1 - x^2] + (x^2 + x^3)*Log[1 - x^2]^2),x]

[Out]

Integrate[(E^((E^x*(-1 + x) + x - x^2 - x*Log[x^2/3] + x*Log[1 - x^2])/(Log[x^2/3] - Log[1 - x^2]))*(2*E^x - 2
*x + (-1 - 2*x - x^2)*Log[x^2/3]^2 + (-x + x^2 + 2*x^3 + E^x*(-x^2 - x^3))*Log[1 - x^2] + (-1 - 2*x - x^2)*Log
[1 - x^2]^2 + Log[x^2/3]*(x - x^2 - 2*x^3 + E^x*(x^2 + x^3) + (2 + 4*x + 2*x^2)*Log[1 - x^2])))/((x^2 + x^3)*L
og[x^2/3]^2 + (-2*x^2 - 2*x^3)*Log[x^2/3]*Log[1 - x^2] + (x^2 + x^3)*Log[1 - x^2]^2), x]

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fricas [A]  time = 0.74, size = 59, normalized size = 1.34 \begin {gather*} \frac {e^{\left (-\frac {x^{2} - {\left (x - 1\right )} e^{x} + x \log \left (\frac {1}{3} \, x^{2}\right ) - x \log \left (-x^{2} + 1\right ) - x}{\log \left (\frac {1}{3} \, x^{2}\right ) - \log \left (-x^{2} + 1\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x-1)*log(1/3*x^2)^2+((2*x^2+4*x+2)*log(-x^2+1)+(x^3+x^2)*exp(x)-2*x^3-x^2+x)*log(1/3*x^2)+(
-x^2-2*x-1)*log(-x^2+1)^2+((-x^3-x^2)*exp(x)+2*x^3+x^2-x)*log(-x^2+1)+2*exp(x)-2*x)*exp((-x*log(1/3*x^2)+x*log
(-x^2+1)+(x-1)*exp(x)-x^2+x)/(log(1/3*x^2)-log(-x^2+1)))/((x^3+x^2)*log(1/3*x^2)^2+(-2*x^3-2*x^2)*log(-x^2+1)*
log(1/3*x^2)+(x^3+x^2)*log(-x^2+1)^2),x, algorithm="fricas")

[Out]

e^(-(x^2 - (x - 1)*e^x + x*log(1/3*x^2) - x*log(-x^2 + 1) - x)/(log(1/3*x^2) - log(-x^2 + 1)))/x

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giac [A]  time = 83.72, size = 43, normalized size = 0.98 \begin {gather*} \frac {e^{\left (-x + \frac {x^{2} - x e^{x} - x + e^{x}}{\log \relax (3) - \log \left (x^{2}\right ) + \log \left (-x^{2} + 1\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x-1)*log(1/3*x^2)^2+((2*x^2+4*x+2)*log(-x^2+1)+(x^3+x^2)*exp(x)-2*x^3-x^2+x)*log(1/3*x^2)+(
-x^2-2*x-1)*log(-x^2+1)^2+((-x^3-x^2)*exp(x)+2*x^3+x^2-x)*log(-x^2+1)+2*exp(x)-2*x)*exp((-x*log(1/3*x^2)+x*log
(-x^2+1)+(x-1)*exp(x)-x^2+x)/(log(1/3*x^2)-log(-x^2+1)))/((x^3+x^2)*log(1/3*x^2)^2+(-2*x^3-2*x^2)*log(-x^2+1)*
log(1/3*x^2)+(x^3+x^2)*log(-x^2+1)^2),x, algorithm="giac")

[Out]

e^(-x + (x^2 - x*e^x - x + e^x)/(log(3) - log(x^2) + log(-x^2 + 1)))/x

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maple [C]  time = 6.40, size = 169, normalized size = 3.84

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 x \ln \relax (3)-4 x \ln \relax (x )+2 x \ln \left (-x^{2}+1\right )+2 \,{\mathrm e}^{x} x -2 x^{2}-2 \,{\mathrm e}^{x}+2 x}{i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 \ln \relax (3)-4 \ln \relax (x )+2 \ln \left (-x^{2}+1\right )}}}{x}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-2*x-1)*ln(1/3*x^2)^2+((2*x^2+4*x+2)*ln(-x^2+1)+(x^3+x^2)*exp(x)-2*x^3-x^2+x)*ln(1/3*x^2)+(-x^2-2*x-
1)*ln(-x^2+1)^2+((-x^3-x^2)*exp(x)+2*x^3+x^2-x)*ln(-x^2+1)+2*exp(x)-2*x)*exp((-x*ln(1/3*x^2)+x*ln(-x^2+1)+(x-1
)*exp(x)-x^2+x)/(ln(1/3*x^2)-ln(-x^2+1)))/((x^3+x^2)*ln(1/3*x^2)^2+(-2*x^3-2*x^2)*ln(-x^2+1)*ln(1/3*x^2)+(x^3+
x^2)*ln(-x^2+1)^2),x,method=_RETURNVERBOSE)

[Out]

1/x*exp(-(I*x*Pi*csgn(I*x^2)^3-2*I*x*Pi*csgn(I*x^2)^2*csgn(I*x)+I*x*Pi*csgn(I*x^2)*csgn(I*x)^2+2*x*ln(3)-4*x*l
n(x)+2*x*ln(-x^2+1)+2*exp(x)*x-2*x^2-2*exp(x)+2*x)/(I*Pi*csgn(I*x^2)^3-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+I*Pi*csg
n(I*x^2)*csgn(I*x)^2+2*ln(3)-4*ln(x)+2*ln(-x^2+1)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left ({\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{3} \, x^{2}\right )^{2} + {\left (x^{2} + 2 \, x + 1\right )} \log \left (-x^{2} + 1\right )^{2} + {\left (2 \, x^{3} + x^{2} - {\left (x^{3} + x^{2}\right )} e^{x} - 2 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (-x^{2} + 1\right ) - x\right )} \log \left (\frac {1}{3} \, x^{2}\right ) - {\left (2 \, x^{3} + x^{2} - {\left (x^{3} + x^{2}\right )} e^{x} - x\right )} \log \left (-x^{2} + 1\right ) + 2 \, x - 2 \, e^{x}\right )} e^{\left (-\frac {x^{2} - {\left (x - 1\right )} e^{x} + x \log \left (\frac {1}{3} \, x^{2}\right ) - x \log \left (-x^{2} + 1\right ) - x}{\log \left (\frac {1}{3} \, x^{2}\right ) - \log \left (-x^{2} + 1\right )}\right )}}{{\left (x^{3} + x^{2}\right )} \log \left (\frac {1}{3} \, x^{2}\right )^{2} - 2 \, {\left (x^{3} + x^{2}\right )} \log \left (\frac {1}{3} \, x^{2}\right ) \log \left (-x^{2} + 1\right ) + {\left (x^{3} + x^{2}\right )} \log \left (-x^{2} + 1\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-2*x-1)*log(1/3*x^2)^2+((2*x^2+4*x+2)*log(-x^2+1)+(x^3+x^2)*exp(x)-2*x^3-x^2+x)*log(1/3*x^2)+(
-x^2-2*x-1)*log(-x^2+1)^2+((-x^3-x^2)*exp(x)+2*x^3+x^2-x)*log(-x^2+1)+2*exp(x)-2*x)*exp((-x*log(1/3*x^2)+x*log
(-x^2+1)+(x-1)*exp(x)-x^2+x)/(log(1/3*x^2)-log(-x^2+1)))/((x^3+x^2)*log(1/3*x^2)^2+(-2*x^3-2*x^2)*log(-x^2+1)*
log(1/3*x^2)+(x^3+x^2)*log(-x^2+1)^2),x, algorithm="maxima")

[Out]

-integrate(((x^2 + 2*x + 1)*log(1/3*x^2)^2 + (x^2 + 2*x + 1)*log(-x^2 + 1)^2 + (2*x^3 + x^2 - (x^3 + x^2)*e^x
- 2*(x^2 + 2*x + 1)*log(-x^2 + 1) - x)*log(1/3*x^2) - (2*x^3 + x^2 - (x^3 + x^2)*e^x - x)*log(-x^2 + 1) + 2*x
- 2*e^x)*e^(-(x^2 - (x - 1)*e^x + x*log(1/3*x^2) - x*log(-x^2 + 1) - x)/(log(1/3*x^2) - log(-x^2 + 1)))/((x^3
+ x^2)*log(1/3*x^2)^2 - 2*(x^3 + x^2)*log(1/3*x^2)*log(-x^2 + 1) + (x^3 + x^2)*log(-x^2 + 1)^2), x)

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mupad [B]  time = 1.02, size = 157, normalized size = 3.57 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \relax (3)}}\,{\mathrm {e}}^{\frac {x^2}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \relax (3)}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \relax (3)}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \relax (3)}}\,{\left (\frac {x^2}{3}\right )}^{\frac {x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \relax (3)}}}{x\,{\left (1-x^2\right )}^{\frac {x}{\ln \left (1-x^2\right )-\ln \left (x^2\right )+\ln \relax (3)}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x + exp(x)*(x - 1) - x*log(x^2/3) - x^2 + x*log(1 - x^2))/(log(1 - x^2) - log(x^2/3)))*(2*x - 2*ex
p(x) - log(x^2/3)*(x + exp(x)*(x^2 + x^3) + log(1 - x^2)*(4*x + 2*x^2 + 2) - x^2 - 2*x^3) + log(x^2/3)^2*(2*x
+ x^2 + 1) + log(1 - x^2)^2*(2*x + x^2 + 1) + log(1 - x^2)*(x + exp(x)*(x^2 + x^3) - x^2 - 2*x^3)))/(log(x^2/3
)^2*(x^2 + x^3) + log(1 - x^2)^2*(x^2 + x^3) - log(1 - x^2)*log(x^2/3)*(2*x^2 + 2*x^3)),x)

[Out]

(exp(-x/(log(1 - x^2) - log(x^2) + log(3)))*exp(x^2/(log(1 - x^2) - log(x^2) + log(3)))*exp(exp(x)/(log(1 - x^
2) - log(x^2) + log(3)))*exp(-(x*exp(x))/(log(1 - x^2) - log(x^2) + log(3)))*(x^2/3)^(x/(log(1 - x^2) - log(x^
2) + log(3))))/(x*(1 - x^2)^(x/(log(1 - x^2) - log(x^2) + log(3))))

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sympy [A]  time = 17.78, size = 44, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {- x^{2} - x \log {\left (\frac {x^{2}}{3} \right )} + x \log {\left (1 - x^{2} \right )} + x + \left (x - 1\right ) e^{x}}{\log {\left (\frac {x^{2}}{3} \right )} - \log {\left (1 - x^{2} \right )}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-2*x-1)*ln(1/3*x**2)**2+((2*x**2+4*x+2)*ln(-x**2+1)+(x**3+x**2)*exp(x)-2*x**3-x**2+x)*ln(1/3*
x**2)+(-x**2-2*x-1)*ln(-x**2+1)**2+((-x**3-x**2)*exp(x)+2*x**3+x**2-x)*ln(-x**2+1)+2*exp(x)-2*x)*exp((-x*ln(1/
3*x**2)+x*ln(-x**2+1)+(x-1)*exp(x)-x**2+x)/(ln(1/3*x**2)-ln(-x**2+1)))/((x**3+x**2)*ln(1/3*x**2)**2+(-2*x**3-2
*x**2)*ln(-x**2+1)*ln(1/3*x**2)+(x**3+x**2)*ln(-x**2+1)**2),x)

[Out]

exp((-x**2 - x*log(x**2/3) + x*log(1 - x**2) + x + (x - 1)*exp(x))/(log(x**2/3) - log(1 - x**2)))/x

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